| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Paired sample confidence interval |
| Difficulty | Standard +0.3 This is a standard paired t-test with confidence interval calculation. Students must compute differences, find mean and standard deviation, then apply textbook formulas for CI and hypothesis test. While it requires careful arithmetic and understanding of paired data, it involves no novel insight—just methodical application of A-level Further Statistics procedures with clearly structured data. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Employee | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) | \(H\) |
| Weight before \(( \mathrm { kg } )\) | 98.6 | 87.3 | 90.4 | 85.2 | 100.5 | 92.4 | 89.9 | 91.3 |
| Weight after \(( \mathrm { kg } )\) | 93.5 | 85.2 | 88.2 | 84.6 | 95.4 | 89.3 | 86.0 | 87.6 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Consider differences e.g.: \(5{\cdot}1\ 2{\cdot}1\ 2{\cdot}2\ 0{\cdot}6\ 5{\cdot}1\ 3{\cdot}1\ 3{\cdot}9\ 3{\cdot}7\) | M1 | |
| Calculate sample mean: \(\bar{d} = 25{\cdot}8/8 = 3{\cdot}225\) | M1 | |
| Estimate population variance: \(s^2 = (100{\cdot}14 - 25{\cdot}8^2/8)/7\) | ||
| (allow biased here: \(2{\cdot}117\) or \(1{\cdot}455^2\)): \(\left[= 2{\cdot}419\ \text{or}\ 1{\cdot}555^2\right]\) | M1 | |
| Find confidence interval (allow \(z\) in place of \(t\)): \(3{\cdot}225 \pm t\sqrt{(2{\cdot}419/8)}\) | M1 | (inconsistent use of 7 or 8 loses M1) |
| Use of correct tabular value: \(t_{7,\ 0{\cdot}975} = 2{\cdot}36[5]\) | A1 | |
| Evaluate C.I. correct to 3 s.f. (in kg): \(3{\cdot}225 \pm 1{\cdot}301\ \text{or}\ [1{\cdot}92,\ 4{\cdot}53]\) | A1 | [Subtotal: 6] |
| State hypotheses: \(H_0: \mu_b - \mu_a = 2{\cdot}5,\ H_1: \mu_b - \mu_a > 2{\cdot}5\) | B1 | |
| Calculate value of \(t\) (to 2 dp): \(t = (\bar{d} - 2{\cdot}5)/(s/\sqrt{8}) = 1{\cdot}32\) | M1 *A1 | |
| Compare with correct tabular \(t\) value: \(t_{7,\ 0{\cdot}95} = 1{\cdot}89[5]\) | *B1 | |
| Correct conclusion (AEF, dep *A1, *B1): Reduction not more than \(2{\cdot}5\) | B1 | [Subtotal: 5] |
## Question 8:
| Working/Answer | Mark | Guidance |
|---|---|---|
| Consider differences e.g.: $5{\cdot}1\ 2{\cdot}1\ 2{\cdot}2\ 0{\cdot}6\ 5{\cdot}1\ 3{\cdot}1\ 3{\cdot}9\ 3{\cdot}7$ | M1 | |
| Calculate sample mean: $\bar{d} = 25{\cdot}8/8 = 3{\cdot}225$ | M1 | |
| Estimate population variance: $s^2 = (100{\cdot}14 - 25{\cdot}8^2/8)/7$ | | |
| (allow biased here: $2{\cdot}117$ or $1{\cdot}455^2$): $\left[= 2{\cdot}419\ \text{or}\ 1{\cdot}555^2\right]$ | M1 | |
| Find confidence interval (allow $z$ in place of $t$): $3{\cdot}225 \pm t\sqrt{(2{\cdot}419/8)}$ | M1 | (inconsistent use of 7 or 8 loses M1) |
| Use of correct tabular value: $t_{7,\ 0{\cdot}975} = 2{\cdot}36[5]$ | A1 | |
| Evaluate C.I. correct to 3 s.f. (in kg): $3{\cdot}225 \pm 1{\cdot}301\ \text{or}\ [1{\cdot}92,\ 4{\cdot}53]$ | A1 | [Subtotal: 6] |
| State hypotheses: $H_0: \mu_b - \mu_a = 2{\cdot}5,\ H_1: \mu_b - \mu_a > 2{\cdot}5$ | B1 | |
| Calculate value of $t$ (to 2 dp): $t = (\bar{d} - 2{\cdot}5)/(s/\sqrt{8}) = 1{\cdot}32$ | M1 *A1 | |
| Compare with correct tabular $t$ value: $t_{7,\ 0{\cdot}95} = 1{\cdot}89[5]$ | *B1 | |
| Correct conclusion (AEF, dep *A1, *B1): Reduction not more than $2{\cdot}5$ | B1 | [Subtotal: 5] |
**Total: 11**8 A company decides that its employees should follow an exercise programme for 30 minutes each day, with the aim that they lose weight and increase productivity. The weights, in kg , of a random sample of 8 employees at the start of the programme and after following the programme for 6 weeks are shown in the table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | }
\hline
Employee & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ \\
\hline
Weight before $( \mathrm { kg } )$ & 98.6 & 87.3 & 90.4 & 85.2 & 100.5 & 92.4 & 89.9 & 91.3 \\
\hline
Weight after $( \mathrm { kg } )$ & 93.5 & 85.2 & 88.2 & 84.6 & 95.4 & 89.3 & 86.0 & 87.6 \\
\hline
\end{tabular}
\end{center}
Assuming that loss in weight is normally distributed, find a 95\% confidence interval for the mean loss in weight of the company's employees.
Test at the $5 \%$ significance level whether, after the exercise programme, there is a reduction of more than 2.5 kg in the population mean weight.
\hfill \mbox{\textit{CAIE FP2 2011 Q8 [11]}}